Week 7 homework help 2015 13 15_19

WEEK 7 HOMEWORK HELP
NUMBERS 13,15 & 19
B Heard
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WEEK 7 HOMEWORK HELP (MORE)
.
They say “it is 60 mg” that means they
are claiming the caffeine content is = 60,
so
H0: µ = 60 (claim)
Ha: µ ≠ 60

.
Remember you are looking at the
Alternative Hypothesis to determine “tails”
Ha: µ ≠ 60 implies we have a two-tailed test

.
As noted in the problem, our alpha is 0.06
Use Minitab to find the critical values
Graph >> Probability Distribution Plots >>
Click View Probability

.
Normal Distribution
Mean of 0
Standard Deviation of 1
Then Click Shaded Area Tab

.
Probability Radial Button
Both Tails
Alpha is 0.06 (The area in the tails)
Input for Probability
Click OK

.
Look at Graph
See values at bottom
They generally round to
Two decimals places
So our z values would be
-1.88 and 1.88
0.4
0.3
0.2
0.1
0.0
X
Density
-1.881
0.03
1.881
0.03
0
Distribution Plot
Normal, Mean=0, StDev=1

.
Find the standardized test statistic (to
compare and see if it is in the critical region)
Minitab
Stat >> Basic Statistics >> 1 Sample z

.
See my inputs and choices based to the
problem
1 – alpha
(100- 6% =
94%)
Alternative was
not equal to

.
See Session Window after clicking ok’s
Standardized Test
Statistic is 1.90

.
See Session Window after clicking ok’s
0.4
0.3
0.2
0.1
0.0
X
Density
-1.881
0.03
1.881
0.03
0
Distribution Plot
1.90 would be barely
in the rejection region
(beyond +1.88), but it
IS in the rejection
region

What do we have?
Remember what we had
H0: µ = 60 (claim)
Ha: µ ≠ 60
Therefore we must REJECT the null that the caffeine
content is 60 and say there is evidence that it is not 60,
so we rejected the null (and the claim in this one) So we
could say that there is enough evidence to reject the
claim.

.
They say “more than 1 pound per person per
day” that means they are claiming it’s more than
1, so
H0: µ ≤ 1
Ha: µ > 1 (claim)

.
Sample size less than 30, so we will use t
distribution

.
We will use Minitab

.
As noted in the problem, the sample size is
20, we will
use Minitab to find the critical t values
Graph >> Probability Distribution Plots >>
Click View Probability

.
t distribution
Degrees of Freedom is
Sample Size Minus 1
20 -1 = 19
Then Click Shaded Area Tab

.
We will reject anything to the right of 1.729
0.4
0.3
0.2
0.1
0.0
X
Density
1.729
0.05
0
Distribution Plot
T, df=19

.
Find the standardized test statistic (to
compare and see if it is in the critical region)
Minitab
Stat >> Basic Statistics >> 1 Sample t

.
Make selections based on problem
1 – alpha
(100- 5% =
95%)
Alternative was
GREATER THAN

Find your standardized test statistic
Here it is,
compare to
my graph

.
It’s to the right of 1.729
0.4
0.3
0.2
0.1
0.0
X
Density
1.729
0.05
0
Distribution Plot
T, df=19
4.47 would be way to
the right, so we are
going to reject the null

.
Remember what we had
H0: µ ≤ 1
Ha: µ > 1 (claim)
By rejecting the null on this one, we are saying that there IS enough
evidence to support the claim (because the claim is the alternative on
this one)

.
They are claiming “more than 20%” or p>0.20, so we set up our null
and alternative
H0: p ≤ 0.20
Ha: µ > 0.20 (claim)
This one screams proportion…. We do three types (1 Sample Z, 1
Sample t and 1 Proportion – this is a “1 Proportion”)

The z test for a proportion is easy, you are just multiplying n times p
and n times q to make sure their products are large enough

We have our null and alternative
H0: p ≤ 0.20

These are similar to 1 sample z when finding the critical values. We
know alpha is equal to 0.03
So in Minitab, go to Graph, Probability Distribution Plots, Click View
Probability, etc.
Normal
Mean of 0
SD of 1

Click Shaded Area Tab, we know alpha is equal to 0.03
Right Tail, because
Ha: µ > 0.20
Alpha

Graph
0.4
0.3
0.2
0.1
0.0
X
Density
1.881
0.03
0
Distribution Plot
We will reject
anything to the right
of this value
(Like a z value they
generally round
these to two decimal
places) So 1.88

Now get your standardized test statistic to compare.
Minitab
Stat >> Basic Statistics >> 1 Proportion

.
In a random sample
of 1150 consumers,
you find 22% have
stopped buying the
product…
(0.22)(1150) = 253
Environmental Agency CLAIMED more than 20% or
0.20
After inputting
everything and
checking Perform
Hypothesis Test,
Click Options
Button

After clicking the options button, change confidence to 97% (1-alpha)
and Alternative to Greater Than AND YOU MUST CHECK BOX THAT
SAYS USE TEST AND INTERVAL BASED ON NORMAL DISTRIBUTION
Greater Than,
because
Ha: µ > 0.20
Alpha = 0.03
1 – 0.03 = .97 or
97%
You MUST Check
this Box

Here is our result including our standardized test statistic
Here it is … 1.70,
now compare to
graph

Graph
0.4
0.3
0.2
0.1
0.0
X
Density
1.881
0.03
0
Distribution Plot
1.70 WOULD NOT BE
IN THE REJECTION
REGION. SO WE DO
NOT REJECT THE
NULL HYPOTHESIS

.
REMEMBER
H0: p ≤ 0.20
By failing to reject the null, we are not supporting the claim… or
saying there is not enough evidence to support the claim (because the
claim is the alternative on this one)

Notice on our results that they also give us the p value…. You could
use either, but they almost ALWAYS specify which they want you to
use…
Here is the p value

This extra stuff was “extra”… Best to you…

Week 7 homework help 2015 13 15_19

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